Question

An object's two dimensional velocity is given by `v(t) = ( 2t^2 , t^3 - 4t )`. What is the object's rate and direction of acceleration at `t=7`?

Answer 1

`|veca(7)|=145.7units`, rounded to one decimal place
Angle with `x`-axis `theta=78.9^@`, rounded to one decimal place

Explanation:

Given two dimensional velocity
`v(t)=(2t^2, t^3-4t)`
Let the two dimensions be `x and y`. Writing explicitly
`vecv(t)=2t^2hati+ (t^3-4t)hat j`

To find acceleration `a(t)`, differentiating with time
`vecv'(t)=veca(t)=d/dt(2t^2hati+ (t^3-4t)hat j)`
or `veca(t)=4thati+ (3t^2-4)hat j`
We need to find acceleration at `t=7`
or `veca(7)=4xx7hati+ (3xx7^2-4)hat j`
`=>veca(7)=28hati+ 143hat j`
Now `|veca(7)|=sqrt(28^2+143^2)`
`|veca(7)|=145.7units`, rounded to one decimal place
If `theta` is the angle acceleration vector makes with `x`-axis
`theta=tan^-1 (143/28)`
`theta=78.9^@`, rounded to one decimal place